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Taylor series of i[x]

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Taylor series of i[x]
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#3
01/13/2017, 05:26 PM
(01/12/2017, 08:50 PM)sheldonison Wrote: I'm not familiar with Sedenion, but it is an abstract algebra concept, not a complex analytic function, right?  So by definition, unless there is some mapping to a complex function, then it would not have a Taylor series...

Well, I believe it is differentialable and if it is, it will have an exact Taylor series.
We just need to get its derivative. I could start to determine it, but I could not finish it, just look at it:
\( di_x / dx = lim (i_{x+h}-i_x)/h = i_x lim (1-i_x / i_{x+h})/h \), where h approaches to 0.
So I know that its derivative will have multiplicative part as i[x] and another thing which is questionable. What is lim (1 - i[x]/i[x+h])/h.
I think it is absolutely solvable ... but how?
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Messages In This Thread
Taylor series of i[x] - by Xorter - 01/12/2017, 04:50 PM
RE: Taylor series of i[x] - by sheldonison - 01/12/2017, 08:50 PM
RE: Taylor series of i[x] - by Xorter - 01/13/2017, 05:26 PM
RE: Taylor series of i[x] - by Xorter - 01/13/2017, 07:13 PM
RE: Taylor series of i[x] - by Xorter - 01/14/2017, 10:14 AM
RE: Taylor series of i[x] - by mike3 - 01/23/2017, 07:38 AM
RE: Taylor series of i[x] - by Xorter - 02/26/2017, 11:10 AM
RE: Taylor series of i[x] - by Xorter - 03/01/2017, 03:06 PM
RE: Taylor series of i[x] - by Xorter - 03/04/2017, 09:40 AM
RE: Taylor series of i[x] - by Xorter - 04/06/2017, 03:42 PM
RE: Taylor series of i[x] - by Xorter - 04/11/2017, 12:18 PM
RE: Taylor series of i[x] - by Xorter - 07/10/2017, 04:07 PM
RE: Taylor series of i[x] - by Xorter - 02/20/2018, 09:55 PM

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